Black Box(POJ 1442·TREAP实现)

传送门：http://poj.org/problem?id=1442

 

Black Box

Time Limit: 1000MS		Memory Limit: 10000K

Description

Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:

ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.

Let us examine a possible sequence of 11 transactions:

Example 1

N Transaction i Black Box contents after transaction Answer 

      (elements are arranged by non-descending)   

1 ADD(3)      0 3   

2 GET         1 3                                    3 

3 ADD(1)      1 1, 3   

4 GET         2 1, 3                                 3 

5 ADD(-4)     2 -4, 1, 3   

6 ADD(2)      2 -4, 1, 2, 3   

7 ADD(8)      2 -4, 1, 2, 3, 8   

8 ADD(-1000)  2 -1000, -4, 1, 2, 3, 8   

9 GET         3 -1000, -4, 1, 2, 3, 8                1 

10 GET        4 -1000, -4, 1, 2, 3, 8                2 

11 ADD(2)     4 -1000, -4, 1, 2, 2, 3, 8   

It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.


Let us describe the sequence of transactions by two integer arrays:


1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).

2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).

The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.

Input

Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.

Output

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

Sample Input

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

Sample Output

3
3
1
2

Source

Northeastern Europe 1996

 

 

第二棵treap。。总体感觉treap非常好写，我学treap是为了防止有的题卡splay。。感觉应该差不多了QAQ...

这题没什么做法。。

Codes:

#include<set>
#include<ctime>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 100100;
#define L(i) (T[i].s[0])
#define R(i) (T[i].s[1])
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)

struct treap{
    int size,s[2],v,pri;
    void Sets(int x,int y){
        size = 1;v = x;pri = y;
    }
}T[N];

int n,m,A[N],size,Lim,now,level = 0;
int tot,root;
int read(){
    char ch = getchar(); int num = 0 , q = 1;
    while(ch>'9'||ch<'0'){
        if(ch=='-') q = -1;
        ch = getchar();
    }
    while(ch>='0'&&ch<='9'){
        num = num * 10 + ch - '0';
        ch = getchar();
    }
    return num * q;
}

void Update(int i){
    T[i].size = T[L(i)].size + T[R(i)].size + 1;
}

void Rot(int &y,int f){
    int x = T[y].s[!f];
    T[y].s[!f] = T[x].s[f];
    T[x].s[f]  = y;
    Update(y);Update(x);
    y = x;
}

void Insert(int &i,int val){
    if(!i){
        T[i=++tot].Sets(val,rand());
        return;
    }
    int f = T[i].v > val;
    Insert(T[i].s[!f],val);
    if(T[T[i].s[!f]].pri > T[i].pri) Rot(i,f);
    else                             Update(i);
}

int Rank(int i,int kth){
    if(T[L(i)].size + 1 == kth) return i;
    else if(T[L(i)].size >=kth) return Rank(L(i),kth);
    else return Rank(R(i),kth - T[L(i)].size - 1);
}

int main(){
    srand(time(NULL));
    n = read(); m = read();
    For(i,n) A[i] = read();
    For(i,m) {
        Lim = read();
        Rep(i,now+1,Lim) Insert(root,A[i]); now = Lim;
        level++;printf("%d\n",T[Rank(root,level)].v);
    }
    return 0;
}